Chapter 12, Problem 20P

To determine

To Calculate: The change in sound level of a sound wave.

Answer to Problem 20P

The sound level of a sound wave is $130\text{dB}$ .

Explanation of Solution

Given:

The given amplitude is, $A=0.13\text{mm}=1.3\times {10}^{-4}\text{m}$

The frequency is, $f=300\text{Hz}$

The sound speed in air, $v=343\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$

The density of the air is, $\rho =1.29\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.$

Formula used:

The intensity is given as,

  $I=2\rho v{\pi }^2{f}^2{A}^2$

Here, $A$ is the amplitude of the wave,

  $\rho$ is the density of the medium,

  $f$ is the frequency of the sound wave,

  $v$ is the speed of the wave.

The intensity is given as,

  $\beta \left(indB\right)=10\log \frac{I}{I_0}$

Here, $I$ is intensity of given sound,

  $I_0$ is the hearing intensity.

Calculation:

By using the given values in the above formula if the intensity is,

  $\begin{array}{l}I=2\rho v{\pi }^2{f}^2{A}^2\\ =2\times \left(1.29\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\right)\times 343\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\times {\left(3.14\right)}^2\times {\left(300\text{Hz}\right)}^2\times {\left(1.3\times {10}^{-4}\text{m}\right)}^2\\ =13.28\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\end{array}$

Now, use this calculated value if the intensity in the sound level formula,

  $\begin{array}{l}\beta \left(\text{in}\text{dB}\right)=10\log \frac{I}{I_0}\\ =10\log \frac{13.28\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.}{1.0\times {10}^{-12}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.} \\ =10\log 13.28\times {10}^{12}\\ =130\text{dB}\end{array}$

Conclusion:

The change in sound level is $130\text{dB}$ .